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san4es73 [151]
3 years ago
13

Help me with these math problems

Mathematics
1 answer:
Tamiku [17]3 years ago
5 0
You cant even see the problems but get photomath it will help u out
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For what values of $x$ is $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$ note: be thorough and explain why all points in your answe
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To solve the inequality \frac{x^2+x+3}{2x^2+x-6}\ge \:0

\mathrm{Factor\:the\:left\:hand\:side\:}\frac{x^2+x+3}{2x^2+x-6}:

The numerator x^2+x+3 is not factorizable.

so factor the denominator 2x^2+x-6:

2x^2+x-6=\left(2x^2-3x\right)+\left(4x-6\right)=x\left(2x-3\right)+2\left(2x-3\right)\\ \mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\ \mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)\\

Now take \mathrm{Factor\:out\:common\:term\:}\left(2x-3\right)

Then we get factor of the denominator as \left(2x-3\right)\left(x+2\right)

Thus \frac{x^2+x+3}{2x^2+x-6}=\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}

Now \mathrm{Compute\:the\:signs\:of\:the\:factors\:of\:}\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}\\

Signs of x^2+x+3>0

\mathrm{Choosing\:ranges\:that\:satisfy\:the\:required\:condition:}\:\ge \:\:0

x\frac{3}{2} is the required solution of the given inequality.

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C; On this line, any two values have to add to 2.

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